Although it is not necessary the meaning might be better expressed: my $n = (@$a > @$b ? @$a : @$b)
Sorry for jumping in -
You cannot compare two arrays that way and expect them to be numerically different - if they are it may be a coincidence; consider for discussion:
#! C:\perl\bin\perl.exe -w #! /usr/bin/perl -w
use strict;
my @aarray=('Start test','Stop test');
my @barray=('ABCDE 1234','T123 S456');my $sanity = (1 > 2) ? "greater" : "lesser"; print "Sanity states: $sanity\n\n";
my $test = (@aarray > @barray) ? ++$#aarray
: ++$#barray;print "Test1 'size' found $test \n\n";
my $test2 = (@barray > @aarray) ? ++$#aarray
: ++$#barray;print "Test2 'size' found $test2 \n\n";
(@aarray <=> @barray) ? print "Array B greater\n"
: print "They are numerically equal\n\n";(@aarray cmp @barray) ? print "Array B is bigger\n"
: print "They are Equal Strings\n\n";__END__
test ? true(1) : false(0);
Comparison <=> cmp 0 if equal, 1 if $a greater, -1 if $b greater
You prolly really meant this:
@a = @b ? @b : @c;See also - perl -le '$a = NaN; print "No NaN support here" if $a == $a' perl -le '$a = NaN; print "NaN support here" if $a != $a'
What you probably want is to compare $#array (array size) or to loop thru each array and test every element (is aarray[1] eq barray[1].)
-Bill- __Sx__________________________________________ http://youve-reached-the.endoftheinternet.org/
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